Calculate electrical power and heating
Use P = VI and its resistance forms to compare power transfer in resistive components.
Section 9 · Electricity
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 9. Circuit calculations and component sketches are original study practice; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use charge, time, energy, power, area and consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation gives power from potential difference and current?
What you will learn
- Recall P = VI, P = I²R and P = V²/R
- Choose a power equation for known circuit quantities
- Interpret electrical power as energy transferred per unit time
Power is energy transferred each second
Electrical power is the rate of energy transfer. Combining W = VQ with I = Q/t gives P = W/t = VI. Power is measured in watts, with 1 W = 1 J s⁻¹.
For an ohmic resistance, substituting V = IR gives P = I²R. Substituting I = V/R gives P = V²/R. These equations are equivalent under the stated resistance model; choose the form matching the known values.
Interpret the comparison conditions
At fixed resistance, increasing current increases resistive heating power in proportion to I². At fixed potential difference, a larger resistance means smaller power V²/R.
Do not use P = I²R or P = V²/R as a universal component law when resistance changes substantially with temperature or operating point. They follow from P = VI and V = IR for the stated resistance at that condition.
Pause and explain
A 4.0 Ω resistor carries 3.0 A. Find its power dissipation.
Worked example
A 12 V component carries 2.0 A. Find power. If it is modelled as a resistor, calculate R and verify with another power equation.
Show the worked solution
- P = VI = 12 × 2.0 = 24 W.
- R = V/I = 12/2.0 = 6.0 Ω.
- P = I²R = (2.0)² × 6.0 = 24 W.
Answer P = 24 W; R = 6.0 Ω; the resistance form agrees.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A lamp operates at 6.0 V and 0.50 A. Find its electrical power at that operating point.
Give me a hint
Use P = VI to multiply the potential difference by the current.
Compare my reasoning
- P = VI.
- P = 6.0 × 0.50.
- State watts.
3.0 W.
Look for these in your work
- Equation selected
- Product correct
- Power unit stated
A resistor has 10 Ω resistance and 2.0 A current. Find the power using P = I²R.
Give me a hint
Square the current in amperes before multiplying by resistance.
Compare my reasoning
- P = (2.0)² × 10.
- P = 4.0 × 10.
- Give the result in watts.
40 W.
Look for these in your work
- Current squared
- Resistance applied
- Watt unit stated
A 24 V heater is modelled as a 12 Ω resistor. Calculate power using V²/R and then determine current and check P = VI.
Give me a hint
Use voltage and resistance first; then use Ohm's law for the check.
Compare my reasoning
- P = V²/R = 24²/12 = 48 W.
- I = V/R = 24/12 = 2.0 A.
- VI = 24 × 2.0 = 48 W.
P = 48 W; I = 2.0 A; P = VI gives the same 48 W.
Look for these in your work
- Suitable formula chosen
- Current checked
- Equivalent power forms agree
Common mistakes
- Confusing power in watts with energy in joules
- Using V²/R with an inconsistent resistance value
- Assuming resistance is fixed for every non-ohmic component
What can you explain now?
Which equation gives power from potential difference and current?
Compare with the explanation
P = VI
Voltage times current is energy transferred per second.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why can the three resistance power equations be used only with a consistent resistance model?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.